［编程之美：3.8］求二叉树中节点的最大距离

//============================================================================
// Name        : maxLenInBtree.cpp
// Author      : qibaoyuan
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================
#include <iostream>
#include <cassert>
using namespace std;
struct LTNOde {
int nMaxLeft;//左子数的最大距离
int nMaxRight;//右子数的最大距离
LTNOde *lChild;
LTNOde *rChild;
int data;
LTNOde() {
nMaxRight = 0;
nMaxLeft = 0;
lChild = NULL;
rChild = NULL;
}
LTNOde(int data) {
nMaxRight = 0;
nMaxLeft = 0;
lChild = NULL;
rChild = NULL;
}
};
//创建一个拥有nNode个节点的完全二叉数
LTNOde *createLBiTree(int nNode) {
LTNOde *root = new LTNOde[nNode];
assert(root);
#define root(i) &root[i-1];
int i;
LTNOde *tNodePtr;
for (i = 0; i < nNode; i++) {
tNodePtr = root(i);
tNodePtr->data = i;
}
for (i = 1; i <= nNode; i++) {
int j = 2 * i;//left child
if (j > nNode)
break;
tNodePtr = root(i);
tNodePtr->lChild = root(j);
if (j < nNode)
tNodePtr->rChild = root(j+1);
}
return root;
}
void findMaxLen(LTNOde *rt, int& nMaxLen) {
if (!rt)
return;//递归出口
if (!rt->lChild)
rt->nMaxLeft = 0;
else {
findMaxLen(rt->lChild, nMaxLen);
rt->nMaxLeft
= rt->lChild->nMaxLeft >= rt->lChild->nMaxRight ? rt->lChild->nMaxLeft
+ 1
: rt->lChild->nMaxRight + 1;
}
if (!rt->rChild)
rt->nMaxRight = 0;
else {
findMaxLen(rt->rChild, nMaxLen);
rt->nMaxRight
= rt->rChild->nMaxLeft >= rt->rChild->nMaxRight ? rt->rChild->nMaxLeft
+ 1
: rt->rChild->nMaxRight + 1;
}
if (rt->nMaxLeft + rt->nMaxRight > nMaxLen)
nMaxLen = rt->nMaxLeft + rt->nMaxRight;
}
int main() {
int const nNode = 3;
LTNOde *biTree = createLBiTree(nNode);
int nMaxLen = 0;
findMaxLen(biTree, nMaxLen);
cout << nMaxLen << endl;
//delete[] biTree;
return 0;
}